Question

`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin 31^circ) - 8 sin^2 30^circ` is equal to ______.

विकल्प

• 1

• – 1

• 0

• 2

Answer 1

`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin 31^circ) - 8 sin^2 30^circ` is equal to 0.

Explanation:

We have, `(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin 31^circ) - 8 sin^2 30^circ`

= `(sin 20^circ)/(sin 20^circ) + (sin 31^circ)/(sin 31^circ) - 8(1/2)^2`  ...[∵ cos θ = sin(90° – θ), sin 30° = 1/2]

= `1 + 1 - 8 xx 1/4`

= 2 – 2

= 0